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3 years ago

11

How does the length and width of a flute affect its sound?

1 answer:

Butoxors [25]3 years ago

4 0

https://www.yamaha.com/en/musical_instrument_guide/flute/mechanism/mechanism002.html

Your welcome!

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How can you make an item that is made of magnetic material become a permanent magnet instead of a temporary magnet?

koban [17]

Answer:

B

Explanation:

You can heat it and then let it cool in a very strong magnetic field.

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What does the sun consume to produce energy and helium?

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This reaction, known as nuclear fusion, converts hydrogen atoms into helium. The by-product of nuclear fusion in theSun's core is a massive volume ofenergy that gets released and radiates outward toward the surface of the Sunand then into the solar system beyond it.

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3 years ago

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Explain why the focal point of a diverging lens is called a virtual focal point.

4vir4ik [10]

The concave lens is a diverging lens, because it causes the light rays to bend away (diverge) from its axis. In this case, the lens has been shaped so that all light rays entering it parallel to its axis appear to originate from the same point, F, defined to be the focal point of a diverging lens.

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3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

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3 years ago

what is the density of a material if its mass is 450 kilograms and occupies a volume of 25 centimeters cubed

asambeis [7]

Answer:

18 kg/cm³

Explanation:Density=mass/ volume=450/25=18kg/cm³

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